CHAPTER 4 – QUANTITIES IN CHEMICAL FORMULAS
4.1 Proportions in Compounds
CO2 vs. CO
 same elements, different proportions = different compounds,
different properties
 CO competes with O2 when binding to haemoglobin in red blood
cells --- if cells don’t get enough O2 --- cell death, tissue
damage, & organ failure-- death
 CO2 ; a) controls acidity of blood
b) solid CO2 = dry ice
c) CO2 gas can be dissolved in water under pressure to
form pop -- when pressure drops bubbles are
released as “gas”
 CO2 also is used in fire extinguishes because doesn’t support
combustion and is denser than air.
LAW OF DEFINITE PROPORTIONS
 says that “ a specific compound always contains the same
elements in definite proportions by mass”-Joseph Proust
 one of first people to apply quantitative analysis to chemistry
----called it stoichiometry
“stoikheion” =means element
“metron” = means measure
 Stoichiometry is the study of the relationships between the
quantities of reactants and products in chemical reactions
4.2 Relative Atomic Mass & Isotopic Abundance
Relative Atomic Mass is the mass of each element needed to react
with a fixed mass of the element chosen to be the standard (carbon-12)
Dalton assigned H an atomic mass of 1, thought water was H:O - 1:1
and thought 1g of H + 8g of O = water…….didn’t know then that water
was H2O or that the ratio would need 16g of O.
 O was chosen as the original standard and was assigned an
atomic mass of 16
 In 1961, carbon-12 was chosen as the standard
 1 atomic mass unit (μ) is defined as exactly 1/12 of the mass of a
C-12 atom i.e. 1 atom of C-12 was defined as having a mass of
12.0μ
Isotopic Abundance –
 is the percentage of an isotope in a sample of an element
 most elements have atomic masses that are not whole numbers,
indicating the presence of isotopes e.g. C-12 & C-13
 naturally occurring samples of carbon consist of 98.89% C-12 and
1.11% C-13
 to calculate atomic mass, we must find the total mass / the
number of atoms:
assume 10,000 atoms, 98.89% C-12 represents 9889 atoms
1.11% C-13 represents 111 atoms
therefore;
mc-12 = 9889 atoms x 12μ =118668μ
mc-13 = 111 atoms x 13μ = 1443μ
mtotal = (118668 + 1443)μ = 120111μ
mav = 120111μ / 10000 atoms = 12.01μ
Try sample problem on p. 165 in your text.
4.3 The Mole and Molar Mass
 ions, atoms and molecules are too small to see so we would need
very large quantities to “see” anything--- instead of using large
numbers we will work with atomic masses and the mole ( a unit of
measurement)
Number
Quantity
Example
2
Pair
Shoes
Dozen
Eggs
144
Gross
Eggs
500
Ream
Paper
6.02 x 1023
MOLE
12
Molecules
AVOGADRO’S CONSTANT – The number of entities in one mole
(NA).
MOLE – The amount of substance containing 6.02 x 1023 entities.
e.g. a) 1 mole of Na atoms is 6.02 x 1023 Na atoms
b) 1 mole of Cl molecules is 6.02 x 1023 Cl molecules
c) 1 mole of NaCl is 6.02 x 1023 formula units of NaCl
d) 1 mole of sheep is 6.02 x 1023 sheep
Molar Mass (M) – the mass of one mole of a substance, expressed
in units of g/mol
 Avogadro’s number is useful in counting atoms & molecules
 scientists know that there are 6.02 x 1023 carbon atoms in 12g of
C-12--- i.e. each atom of C-12 has a mass of 12μ and each mole
of C-12 atoms has a mass of 12g
 the mole makes it easy to convert atomic mass into mass (in
grams)
 e.g. Mg-24….atomic mass is 24μ (2x mass of C-12 which is 12μ
 since a mole of C-12 has a mass of 12g a mole of Mg-24 would
have 2x the mass of C-12 therefore its mass would be 24g
 also applies to molecules, ions, etc.
APPLICATION: atomic mass of H2O = (2 x 1.01μ) + (1 x 16.00μ) =
18.02μ therefore 1 mole of H2O has a mass of 18.02g
BE CAREFUL: when working with molar mass always explain exactly
what you are measuring e.g.
i) molar mass of an oxygen atom = 16.00g/mol but
ii) molar mass of an oxygen molecule = 32.00g/mol
To Calculate Molar Mass (M)
1. Write the chemical formula.
2. Determine the number of atoms (or ions, etc) of each element.
3. Use the atomic mass from the periodic table and the amounts in
moles to determine the molar mass.
4. Communicate the molar mass in the correct units.
Sample Problem:
Q: What is the molecular mass of Ammonium Phosphate, (NH4)3PO4?
A: M(NH4)3 PO4 = 3(N +4H) +P+ 4O
= 3N + 12H P +4O
= (3 x 14.01) + (12 x 1.01) + (1 x 30.97) + (4 x 16.00)
= 149.12g/mol
Converting Mass to Amount in Moles
Amount in moles = m (g)
(# of moles, n)
M (molar mass)
where n= number of moles
m=mass
M=molar mass ------- Or simply n=m/M
e.g. Q: If we have 24g of carbon atoms, how many moles of carbon do
we have?
A: m=24g, M=12g/mol
n=m/M
=24g/12g/mol
=2.0 moles
Sample Problem: Q: Convert a mass of 1.5kg of Calcium Carbonate to
an amount in moles.
A: n=? , m=1.5kg=1500g, M=100.09g/mol
n=m/M
=1500g/100.09g/mol
=15mol
 therefore a mass of 1.5kg of Calcium Carbonate is equal to 15 mol
of Calcium Carbonate.
Converting Amount In Moles to Mass
By manipulating the formula we find that m=n x M.
Sample Problem 2: Q: Convert a reacting amount of 0.346 mol of
Sodium Sulphate into mass (in grams). (see Page 173)
A: m=? , n= 0.346 mol , M=142.04g/mol
m= n x M
=0.346 mol x 142.04g/mol
=49.1g , therefore the mass of 0.346 mol of
Sodium Sulphate is 49.1 g.
Calculations Involving the Number of Entities
N=n x NA
where N=# of entities
n=# of mol
NA=Avogadro’s Constant (6.02 x 1023)
Sample Problem 3: Q: Determine the number of Chloride ions in 0.563
mol of Calcium Chloride (CaCl2)
A: n=0.563
NA=6.02 x 1023
N=n x NA
=0.563 x 6.02 x 1023/mol
=3.39 x 1023
Therefore there are 3.39 formula units in 0.563 mol of CaCl2 BUT
There are 2 chloride ions in CaCl2 so there would actually be 2 x (3.39 x
1023 or 6.78 x 1023 Cl-1 in 0.563 mol of CaCl2
Sample Problem 4: Q: How many sugar (sucrose,C12H22O11(s))
molecules are there in a 1.000kg bag of sugar?
A: M (C12H22O11) = 342.34g/mol
m = 1.000kg = 1000g
n = m/M
= 1000.0g/342.34g/mol
= 2.9211 mol
but using N=n x NA
= 2.9311mol x 6.02 x 1023/mol
= 1.76 x 1024
there are 1.76 x 1024 sugar molecules in a 1.000kg bag of sugar.
Sample Problem 5: Q: What is the mass of 1 water molecule
A: M (H2O) = 18.02g/mol
but because there are 6.02 x 1023 molecules/mol
m (H2O) = 18.02 g/mol x 1mol/6.02 x 1023
= 2.99 x 10-23 g
The mass of 1 water molecule is 2.99 x 10-23 g
Twin Peaks
where m=mass
M=molar mass
N=# of particles
NA =Avogadro’s number
N=# of moles
SECTION 4.5 PERCENTAGE COMPOSITION


to figure out percentage composition, we must compare the mass of
each component in a chemical formula to the total mass of that
chemical
e.g. Q: 2.5g of Hydrogen, when completely reacted, produced 22.5g
of water. What is the percentage composition of water by mass?
A: mH =2.5g
therefore the mO=22.5g – 2.5g =20.0g
MH2O=22.5g
%H=mH/mH2O x 100%
=2.5g/22.5g x 100%=11.1%
%O=mO/mH2O x 100%
=20.0g/22.5g x 100%=88.9%
Sample Problem 1
Q: In an experiment, 3.45g of sodium metal reacted with 5.33g of
chlorine gas to give 8.78g of sodium chloride. Calculate the percentage
composition by mass of sodium chloride.
A: mNa+=3.45g
mCl-=5.33g
mNaCl=8.78g
%Na+=3.45g/8.78g x 100%
=39.3%
%Cl-=5.33g/8.78g x 100%
=60.7%
Therefore the percentage composition by mass of sodium chloride is
39.3% Na and 60.7% Cl.
Percentage Composition Calculations From A Chemical Formula


may be done to establish purity of a compound i.e. pharmaceutical
use
similar to previous problems except that we use atomic mass to
determine the percentages.
Sample Problem 2
Q: Determine the percentage composition of sodium carbonate,
Na2CO3. also known as soda ash
A: mNa=22.99 x 2=45.95
mC=12.01 x 1=12.01
mO=16.00 x 3=48.00
mNa2CO3=105.99
%Na=45.98/105.99 x 100%=43.38%
%C=12.01/105.99 x 100%=11.33%
%O=48.00/105.99 x 100%=45.29%
Therefore, the percentage composition of Na2CO3 is 43.38% sodium,
11.33% carbon and 45.29% oxygen.
SECTION 4.6 EMPIRICAL & MOLECULAR FORMULAS
EMPIRICAL FORMULA – derived from observations in an experiment
rather than from theory
- tells us the simplest ratio of the combining
elements
 combustion analyzer is used to determine percentage composition
of any substance by burning the substance and capturing H2O and
CO2 in chemical traps – used currently for new organic substances
(mostly H and O)
 sometimes 2 or more molecules may have the same percentage
composition but may contain different numbers of atoms in a
molecule e.g. ethyne (acetylene) and benzene have the same
empirical formulas (CH) but different molecular formulas: ethyne
C2H2(g) and benzene-C6H6(l) (see table 1-p.186)
MOLECULAR FORMULA – needed to give the actual number and kind of
atoms in a molecule of the substance
 Definition: A group of chemical symbols used to represent the
number and kind of atoms covalently bonded to form a single
molecule.
SECTION 4.7 CALCULATING CHEMICAL FORMULAS
Calculating an Empirical Formula:
1. Find the mass of each element in 100g of the compound, using
percentage composition.
2. Find the amount in moles of each element by converting the mass
in 100g to moles, using the molar mass of the element.
3. Find the whole-number ratio of atoms in 100g to determine the
empirical formula. Reduce the ratio to the lowest numbers
possible.
Sample Problem 1
Q: What is the empirical formula for a compound whose percentage
composition is 21.6% Na, 33.3% Cl, and 45.1% O?
A: Assume 100g of compound:
Step 1: mNa=21.6% x 100g=21.6g
mCl =33.3% x 100g=33.3g
mO =45.1% x 100g=45.1g
MNa=22.99g/mol
MCl=35.45g/mol
MO=16.00g/mol
Step 2: nNa=21.6g/22.99g/mol
=0.940 mol
nCl=33.3g/35.45g/mol
=0.939mol
nO=45.1g/16.00g/mol
=2.82mol
Step 3: therefore the mole ratio (Na:Cl:O) is 0.940:0.939:2.82
-if we divide the by the smallest number to get the lowest ratio, we get
1:1:3, so therefore the empirical formula is NaClO3.
Calculating A Molecular Formula:
 empirical formula tells us the ratio of atoms in a molecule but not
the number of atoms in a molecule of the compound
 the molar mass takes into account all of the atoms in a molecule
 so, if we know the molar mass of a compound and its empirical
formula, we can determine its molecular formula.
Sample Problem 2:
Q: A compound with a molar mass of 30.00g/mol has an empirical
formula of CH3. Determine its molecular formula.
A: MCH3 =15.04g/mol
If the compound has a molar mass of 30.00g/mol it has twice the
molar mass of the empirical formula, CH3.
Therefore the molecular formula of a compound with a molar mass
of 30.00g is C2H6.
Working With New Compounds
 the problem of working with new compounds is that the molecular
formula may be unknown so we can’t calculate molar masses
using atomic masses from the periodic table
 therefore to determine molar mass we would use a mass
spectrometer.
Sample Problem 3:
Q: What is the molecular formula of the fluid in a butane lighter?
A: Evidence from combustion analysis:
%C=82.5%
%H=17.5%
Evidence from mass spectrometry:
MBUTANE=58g/mol
 for ease of calculation work with one mole (58.0g) of the butane
Step 1: Find mass of each element
mC=82.5/100 x 58.0g=47.8g
mH=17.5/100 x 58.0g=10.2g
Step 2: Convert the mass of each element to # of moles.
nC=47.8g/12.01g/mol
=3.98mol
nH=10.2g/1.01g/mol
=10.0mol
Step 3: The ratio is 4:10 of C:H
Therefore the molecular formula of the fluid in a butane lighter is C4H10.
SUMMARY:
HOW TO DETERMINE A MOLECULAR FORMULA:
A-Given the Empirical Formula & a Measured Molar Mass
1. Calculate the molar mass of the empirical formula.
2. Compare the measured molar mass of the substance with the
molar mass of the empirical formula and increase the subscripts in
the empirical formula by the multiple needed to make the 2 molar
masses equal.
B-Given the Percentage Composition and a Measured Molar Mass
1. Find the mass of each element in one mole of the compound by
multiplying the percentage by the molar mass of the compound.
2. Use the molar mass of the element to convert the mass of the
element to the number of moles.
3. The mole ratio of the elements in the compound provides the
subscripts to use in the molecular formula.